Answer
Verified471.6k+ views
Hint: Probability is a measure of the likelihood that an event will happen.
\[Probability{\text{ }} = \;\dfrac{{favorable\;outcomes}}{{possible\;outcomes}}\]
A conditional probability is the probability of one event if another event occurred. In the “die-toss” example, the probability of event A, three dots showing, is P (A) = \[\dfrac{1}{6}\] on a single toss. But what if we know that event B, at least three dots showing, occurred? Then there are only four possible outcomes, one of which is A. The probability of A = {3} is\[\dfrac{1}{4}\], given that B = {3, 4, 5, 6} occurred. The conditional probability of A given B is written P $\left( {\dfrac{A}{B}} \right)$.
The formula of conditional probability is:
$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A\bigcap B } \right)}}{{P(B)}}$
Event A is independent of B if the conditional probability of A given B is the same as the unconditional probability of A.
Complete step-by-step answer:
A black and a red dice are rolled.
Total possible outcomes =${6^2} = 36$………………… (1)
Let the A be an event obtaining a sum 8
$
\Rightarrow A = \left\{ {\left( {2,6} \right),\left( {3,5} \right),\left( {5,3} \right),\left( {4,4} \right),\left( {6,2} \right)} \right\} \\
\Rightarrow n\left( A \right) = 5 \\
$……………………… (2)
B be an event that the red die resulted in a number less than 4.
\[
\Rightarrow B =
\left\{
\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right), \\
\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right), \\
\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right), \\
\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right), \\
\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right), \\
\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right) \\
\right\} \\
\Rightarrow n\left( B \right) = 18 \\
\] ……………………………….(3)
Let’s find intersect when both event A and B happens by equation (2) and (3):-
$
\Rightarrow A\bigcap B = \left\{ {\left( {5,3} \right),(6,2)} \right\} \\
\Rightarrow n\left( {A\bigcap B } \right) = 2 \\
$
Probability of $A\bigcap B $= \[\dfrac{{n\left( {A\bigcap B } \right)}}{{total\;possible\;outcomes}} = \dfrac{2}{{36}}..................(1)\]
Probability of event B = P (B) =$\dfrac{{18}}{{36}}$………………………….(2)
Using the formula mentioned in the hint of conditional probability and putting the values of equation (1) and (2).
$
P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A\bigcap B } \right)}}{{P(B)}} \\
\Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{{\dfrac{2}{{36}}}}{{\dfrac{{18}}{{36}}}} \\
\Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{9} \\
$
The conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4 is $\dfrac{1}{9}$
Note: Conditional probability is used in many areas, in fields as diverse as calculus, insurance, and politics. For example, the re-election of a president depends upon the voting preference of voters and perhaps the success of television advertising—even the probability of the opponent making gaffes during debates!
\[Probability{\text{ }} = \;\dfrac{{favorable\;outcomes}}{{possible\;outcomes}}\]
A conditional probability is the probability of one event if another event occurred. In the “die-toss” example, the probability of event A, three dots showing, is P (A) = \[\dfrac{1}{6}\] on a single toss. But what if we know that event B, at least three dots showing, occurred? Then there are only four possible outcomes, one of which is A. The probability of A = {3} is\[\dfrac{1}{4}\], given that B = {3, 4, 5, 6} occurred. The conditional probability of A given B is written P $\left( {\dfrac{A}{B}} \right)$.
The formula of conditional probability is:
$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A\bigcap B } \right)}}{{P(B)}}$
Event A is independent of B if the conditional probability of A given B is the same as the unconditional probability of A.
Complete step-by-step answer:
A black and a red dice are rolled.
Total possible outcomes =${6^2} = 36$………………… (1)
Let the A be an event obtaining a sum 8
$
\Rightarrow A = \left\{ {\left( {2,6} \right),\left( {3,5} \right),\left( {5,3} \right),\left( {4,4} \right),\left( {6,2} \right)} \right\} \\
\Rightarrow n\left( A \right) = 5 \\
$……………………… (2)
B be an event that the red die resulted in a number less than 4.
\[
\Rightarrow B =
\left\{
\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right), \\
\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right), \\
\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right), \\
\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right), \\
\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right), \\
\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right) \\
\right\} \\
\Rightarrow n\left( B \right) = 18 \\
\] ……………………………….(3)
Let’s find intersect when both event A and B happens by equation (2) and (3):-
$
\Rightarrow A\bigcap B = \left\{ {\left( {5,3} \right),(6,2)} \right\} \\
\Rightarrow n\left( {A\bigcap B } \right) = 2 \\
$
Probability of $A\bigcap B $= \[\dfrac{{n\left( {A\bigcap B } \right)}}{{total\;possible\;outcomes}} = \dfrac{2}{{36}}..................(1)\]
Probability of event B = P (B) =$\dfrac{{18}}{{36}}$………………………….(2)
Using the formula mentioned in the hint of conditional probability and putting the values of equation (1) and (2).
$
P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A\bigcap B } \right)}}{{P(B)}} \\
\Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{{\dfrac{2}{{36}}}}{{\dfrac{{18}}{{36}}}} \\
\Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{9} \\
$
The conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4 is $\dfrac{1}{9}$
Note: Conditional probability is used in many areas, in fields as diverse as calculus, insurance, and politics. For example, the re-election of a president depends upon the voting preference of voters and perhaps the success of television advertising—even the probability of the opponent making gaffes during debates!